我需要在输入时替换EditText中的文本: 示例:如果用户按下“A”,它将被存储到缓冲区中并且在EditText上显示“D”(看起来他按下了“D”)。 现在我可以读取按下的字符但我不能在et中显示任何字符以避免stackoverflow:
final EditText et = (EditText) findViewById(R.id.editTexts);
final TextView tv = (TextView) findViewById(R.id.textView2);
et.addTextChangedListener(new TextWatcher()
{
public void afterTextChanged(Editable s){}
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
}
public void onTextChanged(CharSequence s, int start, int before, int count) {
if(s.length() > 0) {
tv.setText(s.toString().substring(s.length()-1));
et.setText("");
}
}
});
答案 0 :(得分:26)
您可以根据需要进行更改::
public class SampleActivity extends Activity {
TextWatcher tt = null;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
final EditText et = (EditText) findViewById(R.id.editText1);
final TextView tv = (TextView) findViewById(R.id.textView1);
tt = new TextWatcher() {
public void afterTextChanged(Editable s){
et.setSelection(s.length());
}
public void beforeTextChanged(CharSequence s,int start,int count, int after){}
public void onTextChanged(CharSequence s, int start, int before, int count) {
et.removeTextChangedListener(tt);
et.setText(et.getText().toString().replace("A", "C"));
et.addTextChangedListener(tt);
}
};
et.addTextChangedListener(tt);
}
}
答案 1 :(得分:8)
要以交互方式更改文本,您需要注册TextWatcher
。但是尝试更改观察者内部的文本会进一步调用观察者。我的黑客是在我想要更改文本时临时删除观察者,并在
mEditText.addTextChangedListener(new TextWatcher() {
@Override public void beforeTextChanged(CharSequence charSequence, int i, int i1, int i2) { }
@Override public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) { }
@Override public void afterTextChanged(Editable editable) {
mEditText.removeTextChangedListener(this);
mEditText.setText(//TODO change whatever you like here);
mEditText.setSelection(editable.length()); //moves the pointer to end
mEditText.addTextChangedListener(this);
}
答案 2 :(得分:2)